RWM102: Algebra

3a. Translate a verbal expression into a variable expression 

• How do you create an expression from a word problem?

When solving word problems, one of the challenges is correctly converting the written words to a mathematical expression. When reading a word problem, try to look for these keywords that often represent mathematical operations:

Sometimes the name of the variable is explicitly given, such as "The product of $x$ and 3 is equal to 7", which means $3x=7$.

Other times, the variable is not named, and you can use any variable you choose, such as "Three more than a number is equal to 10", which means $x+3=10$.

To review, see Translating Words into Mathematical Symbols. 

3b. Use the basic percent equation to solve problems involving percents 

• How do you solve percent problems?
• How do you convert a percent to a decimal?
• How do you convert a decimal to a percent?

When working with percents, it is important to remember that the word percent means "out of 100". Therefore, we often multiply or divide by 100.

When converting a decimal to a percent, we must multiply by 100. For example, .345 is multiplied by 100 to give 34.5%.

On the other hand, sometimes we have a percent and we want to convert it to a decimal. This is done by dividing by 100. For example, 45% is divided by 100, giving a decimal value of .45.

Similarly, sometimes a fraction needs to be converted to a percent. To do this, simply divide the fraction to get a decimal, then convert the decimal to a percent as we just did.

When solving percent problems, it is helpful to use a standard percent equation, such as the equation:

$part = \% rate \times base$

For example, if you are asked to find 45% of 50, use the percent equation to find $part=45 \% \times 50$. Convert the percent to a decimal, and then multiply to find $part= .45 \times 50=22.5$.

To review, see Solving Percent Word Problems and Practice Solving Percent Word Problems.

3c. Apply the basic percent equation to problems involving mixtures, markups, and discounts 

• How do you solve percent problems with mixtures?
• How do you solve percent problems with markups and discounts?

A common percent problem is the mixture problem. In these problems, you are trying to combine two different liquids of different strengths to make a mixture of a new strength. For example, if you have 20L of 25% sugar solution, how much of a 10% sugar solution would you have to mix to end up with a 15% sugar solution?

This problem is just the percent equation used 3 times together. Set up the problem with the percent strength times the volume for each of the two mixtures you have, and remember that will equal the total amount of 15% solution in the end. The amount of the 10% solution you need to add should be x.

$.25 \times 20+.10x=.15(x+20)$

By simplifying, using the distributive property, and solving, we will find that $x = 40L$.

When calculating markups or discounts, we can use the percent change equation, which is:

$\% change=\dfrac{final \; amount - original \; amount}{original \; amount} \times 100$

For example, if a dress has an original price of $80, and is on sale for $50, you can calculate the discount by substituting the values into the percent change equation.

$\% change=\dfrac{50-80}{80} \times 100= -37.5 \%$

This means we have a 37.5% discount. If the price was increased, instead of decreased, the percent change equation will calculate the markup, instead of the discount.

To review, see:

• Mixture Problems
• Solving Mixture Problems
• Practice Solving Mixture Problems
• Solving Percent Word Problems

3d. Use the uniform motion equation to solve problems involving uniform motion

• How do you solve problems involving distance, time, and speed?

When solving problems involving the time it takes to travel a certain distance, you should use the uniform motion equation, $d=rt$, where $d$ is the distance, $r$ is the rate (speed), and t is the time. When solving a uniform motion problem, 2 of the 3 variables will be given, you will need to find the third unknown variable.

For example, if Cindy wants to travel to see her grandmother 240 miles away and she drives at 60 miles per hour, how long will it take her to reach her grandmother's house? In this problem, 240 miles is the distance, and 60 miles per hour is the rate, so if we substitute into the uniform motion equation, we get \)240=60t\). Now, we only need to divide both sides by 60 to find $t = 4$. So, it will take her 4 hours to reach her grandmother's house.

To review, see Solving a Formula for a Specific Value. 

3e. Create equations in one variable and use them to solve problems 

• How do you create an equation from a word problem?
• How do you solve one-variable word problems?

When solving word problems, it is often necessary to create a mathematical equation from the information given to solve the problem. When creating an equation for a word problem, you can follow these steps:

1. Let $x$ (or some other letter) represent the unknown quantity.
2. Translate the words to mathematical symbols and form an equation. Draw a picture if possible.
3. Solve the equation.
4. Check the solution by substituting the result into the original statement, not equation, of the problem.
5. Write a conclusion.

For example, consider the question "when 3 times a number is increased by 4, the result is eight more than 5 times the number". In this case, we can translate the words into the equation $3x+4=5x+8$. Using our skills for solving equations, you will find that $x=-2$.

To review, see Solving Algebraic Expressions and Equations Part I.

Unit 3 Vocabulary

This vocabulary list includes terms listed above that students need to know to successfully complete the final exam for the course.

• converting 
• mathematical expression 
• percent
• percent equation
• mixture problem
• markups
• discounts 
• percent change equation
• uniform motion equation
• word problems